# Homework Help: Abs max/min

1. Jul 20, 2008

### fk378

1. The problem statement, all variables and given/known data
Find the absolute max and min values of f on the set D.

f(x,y)=4xy^3 - (x^2)(y^2) - xy^3
D is the closed triangular region in the xy-plane with vertices (0,0) (0,6) and (6,0).

3. The attempt at a solution
I found my two critical points to be (1,2) and (2,0). Then I tried to evaluate the boundary points:
1) 0<x<6, y=0
2) 0<y<6, x=0
3) (6-y, y) because the third boundary line is y= -x+6

I don't know how to solve for the last boundary line though. I plugged in x=6-y in the original equation, got the expression (2y^3)-(12y^2). Do I just plug in numbers now?

My book gets (2,4) for the absolute min, which is a point on this 3rd boundary line. I just don't see how to come up with the point, though.

2. Jul 20, 2008

### HallsofIvy

Did you write this correctly? Why not just 3xy3- x2y2?

f(x, 0)= 0 so that's easy.

f(0,y)= 0 so that's easy.

If what you wrote before is correct, f(6-y,y) will be a fourth degree polynomial, not a cubic. In any case, what every you get as a function of y alone, with y between 0 and 6, you differentiate and set equal to 0 to find the max or min.

Last edited by a moderator: Jul 21, 2008
3. Jul 20, 2008

### fk378

Sorry! It is a typo. It should read:
Find the absolute max and min values of f on the set D.

f(x,y)=4xy^2 - (x^2)(y^2) - xy^3

So, if the other values of f(0,y) and f(x,0) did not give values of 0, would I also need to differentiate their functions and find the critical points there as well?